Why we define ''f'' with between -fs/2 and fs/2.
fs is the sampling frequency.
f is the signal frequency and for successful reconstruction, sampling frequency must be equal to or greater than twice the signal frequency. ( According to Sampling Theorem).
fs= 2*f implies f=fs/2.
For both negative and positive time period of signal, f is defined between - fs/2 and +fs/2.
Why we define ''f'' with between -fs/2 and fs/2. %fft DSB modulation; ts-1/fs tmax-(N-1)*ts; t-0:ts:tmax f--fs/2:fs/(N-1):fs/2; y2-fftshiftlftlv subplot(10,1,6) plot(f,y2) title('...
Amplitude=3; fs=8000; n=0:399; t=0:1/fs: n*1/fs-1/fs; signal=3+3*cos(2*pi*1100*t)+3*cos(2*pi*2200*t)+3*cos(2*pi*3300*t); fftSignal= fft(signal); fftSignal=f ftshift (fftSignal); f=fs/2*linspace(-1,1,fs); plot(f,abs(fftsignal); xlabel('Frequency(Hz)’) ylabel('amplitude(v)') title('Spectral domain') plz code above using For ..End loop to archive the same results.
Can you please help me answer Task 2.b? Please show all work. fs=44100; no_pts=8192; t=([0:no_pts-1]')/fs; y1=sin(2*pi*1000*t); figure; plot(t,y1); xlabel('t (second)') ylabel('y(t)') axis([0,.004,-1.2,1.2]) % constrain axis so you can actually see the wave sound(y1,fs); % play sound using windows driver. %% % Check the frequency domain signal. fr is the frequency vector and f1 is the magnitude of F{y1}. fr=([0:no_pts-1]')/no_pts*fs; %in Hz fr=fr(1:no_pts/2); % single-sided spectrum f1=abs(fft(y1)); % compute fft f1=f1(1:no_pts/2)/fs; %% % F is the continuous time Fourier. (See derivation...
1. (9 points) In this Question, we are going to perform DSBSC modulation using MAT- LAB. The signal we want to use is a speech signal. Here is the block diagram of ths system we want to simulate: Modulation Demodulation ult gt) m(t) x Butterworth LPF mr(t) c(t) Gr(t) Figure 1: DSBSC modulation and demodulation. (a) Since we are working with speech signals, we will choose a sampling frequency that is much larger than the bandwidth of the signals. As...
Exercises: u used to the instructor b the end of next lab. 20 102 Plot the f(t)-sinc(20r) cos(300t)sinc (10t) cos(100t) Use the fast Fourier transform to find the magnitude and phase spectrum of the signal and plot over an appropriate range. Use appropriate values for the time interval and the sampling interval. Note that in Matlab sinc(x)-, so we need to divide the argument by n to make it match the given function. Le, sinc(20t/pi) Hint: Use the parameters from...
Program from problem 1: (Using MATLAB) % Sampling frequency and sampling period fs = 10000; ts = 1/fs; % Number of samples, assume 1000 samples l = 1000; t = 0:1:l-1; t = t.*ts; % Convert the sample index into time for generation and plotting of signal % Frequency and amplitude of the sensor f1 = 110; a1 = 1.0; % Frequency and amplitude of the power grid noise f2 = 60; a2 = 0.7; % Generating the sinusoidal waves...
Question 2 [30pts]: By using a 3x1 subplot, plot x(t) signal in the first row. Take t between 0 s and 10 s with an increment of 1 ms. Furthermore, calculate and plot the Fourier series representation of x(t) = Ek--Nakejkwot for N = 10 and N = 100 in the second and third rows of the subplot. Note that, x(t) = Na elkwot should be calculated by MATLAB. Please comment on increasing N on your results. Do not forget...
MATLAB code starts here --------- clear T0=2; w0=2*pi/T0; f0=1/T0; Tmax=4; Nmax=15; %--- i=1; for t=-Tmax: .01:Tmax T(i)=t; if t>=(T0/2) while (t>T0/2) t=t-T0; end elseif t<=-(T0/2) while (t<=-T0/2) t=t+T0; end end if abs(t)<=(T0/4) y(i)=1; else y(i)=0; end i=i+1; end plot(T,y),grid, xlabel('Time (sec)'); title('y(t) square wave'); shg disp('Hit return..'); pause %--- a0=1/2; F(1)=0; %dc freq C(1)=a0; for n=1:Nmax a(n)=(2/(n*pi))*sin((n*pi)/2); b(n)=0; C(n+1)=sqrt(a(n)^2+b(n)^2); F(n+1)=n*f0; end stem(F,abs,(C)), grid, title(['Line Spectrum: Harmonics = ' num2str(Nmax)]); xlabel('Freq(Hz)'), ylabel('Cn'), shg disp('Hit return...'); pause %--- yest=a0*ones(1,length(T)); for n=1:Nmax yest=yest+a(n)*cos(2*n*pi*T/T0)+b(n)*sin(2*n*pi*T/T0);...
I wrote a Matlab program for the figure below. When I plot the waves, they look the same. Why do the two waves frequencies look same and How do I avoid it? (I really do need this part of the question answered.) N = 200; % Total number of time domain samples in simulation. Fs = 100 ;% sampling frequency. F1 = 10; % frequency of wave - 1. F2 = 90; % frequency of wave - 2. phi =...
Create a file named “toneA.m” with the following MATLAB code: clear all Fs=5000; Ts=1/Fs; t=[0:Ts:0.4]; F_A=440; %Frequency of note A is 440 Hz A=sin(2*pi*F_A*t); sound(A,Fs); Type toneA at the command line and then answer the following: (a) What is the time duration of A? (b) How many elements are there in A? (c) Modify toneA.m by changing “Fs=5000” to “Fs=10000”. Can you hear any difference? (d) Create a file named “tone.m” with the following MATLAB code: function x = tone(frequency,...
NB! This task is required to be solved in matlab. this task also requires the use of the function displayDualSpectrum(); which i have pasted in the bottom. the tasks that i need help with are A), B) and C). this is a multi-part question. Task - Frequency mixing We use a basic signal that can be described mathematically as follows: with this We shall then make an amplitude modulated signal: where fc is the carrier frequency. the code below specifies...